[proofplan] We reduce to the nonneg case via the Jordan decomposition. By the [Jordan Decomposition Theorem](/theorems/1212), $\nu = \nu^+ - \nu^-$ with $\nu^+, \nu^- \ge 0$ and $\nu^+ \perp \nu^-$. The condition $|\nu| \ll \mu$ implies $\nu^+ \ll \mu$ and $\nu^- \ll \mu$. Applying the [Radon-Nikodym Theorem](/theorems/1247) to each nonneg part and taking the difference yields the density $f = f^+ - f^-$. The identity $|f| = \frac{d|\nu|}{d\mu}$ follows from the relation $|\nu| = \nu^+ + \nu^-$ and the [Linearity of the Radon-Nikodym Derivative](/theorems/1210). [/proofplan] [step:Apply the Jordan Decomposition to obtain $\nu = \nu^+ - \nu^-$] By the [Jordan Decomposition Theorem](/theorems/1212), there exist unique nonneg measures $\nu^+$ and $\nu^-$ on $(X, \mathcal{A})$ with $\nu^+ \perp \nu^-$ and $\nu = \nu^+ - \nu^-$. The total variation measure satisfies $|\nu| = \nu^+ + \nu^-$. Both $\nu^+$ and $\nu^-$ are $\sigma$-finite. Since $\nu$ is a $\sigma$-finite signed measure, there exists an exhaustion $(X_k)_{k=1}^\infty$ with $X_k \uparrow X$ and $|\nu|(X_k) < \infty$. Then $\nu^+(X_k) \le |\nu|(X_k) < \infty$ and $\nu^-(X_k) \le |\nu|(X_k) < \infty$ for each $k$, so both $\nu^+$ and $\nu^-$ are $\sigma$-finite. [guided] The Jordan Decomposition is the signed-measure analogue of writing a real number as the difference of its positive and negative parts. The decomposition $\nu = \nu^+ - \nu^-$ is constructed from a Hahn decomposition $X = P \cup N$: set $\nu^+(A) = \nu(A \cap P)$ and $\nu^-(A) = -\nu(A \cap N)$. The total variation $|\nu| = \nu^+ + \nu^-$ is a nonneg measure that dominates both $\nu^+$ and $\nu^-$. The condition $|\nu| \ll \mu$ in the theorem statement is therefore a natural requirement: it ensures both components are absolutely continuous with respect to $\mu$. [/guided] [/step] [step:Verify $\nu^+ \ll \mu$ and $\nu^- \ll \mu$ from $|\nu| \ll \mu$] Let $A \in \mathcal{A}$ with $\mu(A) = 0$. Since $|\nu| \ll \mu$, we have $|\nu|(A) = 0$. Since $\nu^+(A) \le |\nu|(A)$ and $\nu^-(A) \le |\nu|(A)$ (both are nonneg measures dominated by $|\nu|$), we conclude $\nu^+(A) = 0$ and $\nu^-(A) = 0$. Therefore $\nu^+ \ll \mu$ and $\nu^- \ll \mu$. [guided] The domination $\nu^+ \le |\nu|$ and $\nu^- \le |\nu|$ follows directly from $|\nu| = \nu^+ + \nu^-$: for any $A \in \mathcal{A}$, $\nu^+(A) = |\nu|(A) - \nu^-(A) \le |\nu|(A)$ since $\nu^-(A) \ge 0$. This is why the hypothesis $|\nu| \ll \mu$ (rather than just $\nu \ll \mu$, which is not well-defined for signed measures without specifying what "null set" means) is the correct condition. [/guided] [/step] [step:Apply the Radon-Nikodym Theorem to $\nu^+$ and $\nu^-$ separately] Since $\nu^+ \ll \mu$ and both $\nu^+$ and $\mu$ are $\sigma$-finite, the [Radon-Nikodym Theorem](/theorems/1247) provides an $\mathcal{A}$-measurable function: \begin{align*} f^+: X \to [0, \infty), \quad \nu^+(A) = \int_A f^+ \, d\mu \quad \text{for every } A \in \mathcal{A}. \end{align*} Similarly, since $\nu^- \ll \mu$ and both are $\sigma$-finite, there exists: \begin{align*} f^-: X \to [0, \infty), \quad \nu^-(A) = \int_A f^- \, d\mu \quad \text{for every } A \in \mathcal{A}. \end{align*} [/step] [step:Define $f := f^+ - f^-$ and verify the integral representation] Define the $\mathcal{A}$-measurable function: \begin{align*} f: X &\to \mathbb{R}, \quad f := f^+ - f^-. \end{align*} We verify $f \in L^1(X, \mathcal{A}, \mu)$. Since $|\nu|$ is $\sigma$-finite and $|\nu| \ll \mu$: \begin{align*} \int_X |f| \, d\mu = \int_X (f^+ + f^-) \, d\mu = \int_X f^+ \, d\mu + \int_X f^- \, d\mu = \nu^+(X) + \nu^-(X) = |\nu|(X). \end{align*} The signed measure $\nu$ is finite (since $|\nu|(X) < \infty$ follows from the definition of a signed measure, which requires at least one of $\nu^+(X)$ and $\nu^-(X)$ to be finite — but here $\sigma$-finiteness of $|\nu|$ combined with the exhaustion gives $|\nu|(X) = \lim_k |\nu|(X_k) \le \infty$). If $|\nu|(X) = \infty$, then $f$ is not globally integrable, but $f$ is locally integrable on each $X_k$ with $|\nu|(X_k) < \infty$, which suffices for the integral representation on each measurable set. For every $A \in \mathcal{A}$: \begin{align*} \int_A f \, d\mu &= \int_A f^+ \, d\mu - \int_A f^- \, d\mu = \nu^+(A) - \nu^-(A) = \nu(A). \end{align*} The subtraction is valid because at least one of $\int_A f^+ \, d\mu = \nu^+(A)$ and $\int_A f^- \, d\mu = \nu^-(A)$ is finite (since $\nu$ is a signed measure, at least one of $\nu^+(A)$, $\nu^-(A)$ is finite for every $A$). [guided] The subtraction $\int_A f^+ \, d\mu - \int_A f^- \, d\mu$ requires justification: we need at least one of the two integrals to be finite to avoid the indeterminate form $\infty - \infty$. This is guaranteed by the definition of a signed measure: $\nu$ takes values in $(-\infty, \infty]$ or $[-\infty, \infty)$, meaning at least one of $\nu^+$ and $\nu^-$ is a finite measure. If $\nu^-$ is finite, then $\int_A f^- \, d\mu = \nu^-(A) \le \nu^-(X) < \infty$, making the subtraction well-defined. The symmetric case holds if $\nu^+$ is finite. [/guided] [/step] [step:Prove uniqueness and the identity $|f| = \frac{d|\nu|}{d\mu}$] **Uniqueness.** Suppose $g: X \to \mathbb{R}$ also satisfies $\nu(A) = \int_A g \, d\mu$ for every $A \in \mathcal{A}$. Then $\int_A (f - g) \, d\mu = 0$ for every $A \in \mathcal{A}$. Taking $A = \{f - g > 0\}$ gives $\int_{\{f-g>0\}} (f-g) \, d\mu = 0$, so $f - g \le 0$ $\mu$-a.e. on $\{f - g > 0\}$, hence $\mu(\{f - g > 0\}) = 0$. Similarly $\mu(\{f - g < 0\}) = 0$. Therefore $f = g$ $\mu$-a.e. **Total variation identity.** The [Linearity of the Radon-Nikodym Derivative](/theorems/1210) gives: \begin{align*} \frac{d|\nu|}{d\mu} = \frac{d(\nu^+ + \nu^-)}{d\mu} = f^+ + f^- = |f^+ - f^-| = |f| \quad \mu\text{-a.e.}, \end{align*} where the third equality uses the fact that $f^+ \cdot f^- = 0$ $\mu$-a.e. [claim:Disjointness of supports: $f^+ \cdot f^- = 0$ $\mu$-a.e.] The functions $f^+ = \frac{d\nu^+}{d\mu}$ and $f^- = \frac{d\nu^-}{d\mu}$ satisfy $f^+(x) \cdot f^-(x) = 0$ for $\mu$-a.e. $x \in X$. [/claim] [proof] Since $\nu^+ \perp \nu^-$, there exists a partition $X = P \cup N$ with $P \cap N = \varnothing$ such that $\nu^+(N) = 0$ and $\nu^-(P) = 0$. For any measurable $A \subseteq N$: \begin{align*} \int_A f^+ \, d\mu = \nu^+(A) \le \nu^+(N) = 0, \end{align*} so $f^+ = 0$ $\mu$-a.e. on $N$. Similarly, $f^- = 0$ $\mu$-a.e. on $P$. Since $X = P \cup N$, the product $f^+ \cdot f^- = 0$ $\mu$-a.e. [/proof] Since $f^+$ and $f^-$ have $\mu$-a.e. disjoint supports, $|f| = |f^+ - f^-| = f^+ + f^-$ $\mu$-a.e. (on the set where $f^+ > 0$ we have $f^- = 0$ so $|f| = f^+ = f^+ + f^-$; on the set where $f^- > 0$ we have $f^+ = 0$ so $|f| = f^- = f^+ + f^-$; on the set where both vanish, $|f| = 0 = f^+ + f^-$). Combined with the linearity identity, $|f| = \frac{d|\nu|}{d\mu}$ $\mu$-a.e. [guided] The identity $|f| = f^+ + f^-$ relies on the disjoint support property, which in turn comes from the mutual singularity $\nu^+ \perp \nu^-$. Without mutual singularity, we would only have $|f| \le f^+ + f^-$. The Jordan Decomposition Theorem guarantees $\nu^+ \perp \nu^-$, which is what makes the total variation formula $|f| = \frac{d|\nu|}{d\mu}$ work. This is the structural reason why the total variation $|\nu|$ is defined as $\nu^+ + \nu^-$ (using the Jordan decomposition) rather than by a supremum formula: the Jordan decomposition provides the cleanest path to the density formula. [/guided] [/step]